Question:
Three positive and two negative charges each of magnitude \(q\) are placed at five of the six vertices of a regular hexagon of side \(R\). The work done to bring a negative charge \(-q\) from infinity to the centre of the hexagon is:
Three positive and two negative charges each of magnitude \(q\) are placed at five of the six vertices of a regular hexagon of side \(R\). The work done to bring a negative charge \(-q\) from infinity to the centre of the hexagon is:
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Potential is a scalar quantity. Therefore, while calculating potential at a point, charges are added algebraically without considering direction.
Updated On: Jun 11, 2026
- \(\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{-q^2}{R}\right)\)
- \(\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q^2}{R}\right)\)
- \(\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{-5q^2}{R}\right)\)
- \(\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q^2}{2R}\right)\)
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The Correct Option is A
Solution and Explanation
Concept:
Work done in bringing a charge from infinity equals the increase in potential energy.
\[
W=q_0V
\]
where \(V\) is the potential at the destination point.
Step 1: Find the net charge contributing to potential. There are: \[ 3(+q)+2(-q) \] Therefore, \[ Q_{\text{net}}=+q \]
Step 2: Determine distance from centre. For a regular hexagon, \[ \text{Circumradius}=R \] Hence every occupied vertex lies at distance \(R\) from the centre.
Step 3: Calculate potential at centre. Potential due to one charge: \[ V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R} \] Adding algebraically: \[ V= \frac{1}{4\pi\varepsilon_0} \frac{(+q)}{R} \]
Step 4: Compute work done. Charge brought: \[ q_0=-q \] Hence \[ W=q_0V \] \[ W=(-q) \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{R} \right) \] \[ W= \frac{1}{4\pi\varepsilon_0} \left( \frac{-q^2}{R} \right) \] Thus, \[ \boxed{ W= \frac{1}{4\pi\varepsilon_0} \left( \frac{-q^2}{R} \right) } \] Therefore option (A) is correct.
Step 1: Find the net charge contributing to potential. There are: \[ 3(+q)+2(-q) \] Therefore, \[ Q_{\text{net}}=+q \]
Step 2: Determine distance from centre. For a regular hexagon, \[ \text{Circumradius}=R \] Hence every occupied vertex lies at distance \(R\) from the centre.
Step 3: Calculate potential at centre. Potential due to one charge: \[ V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R} \] Adding algebraically: \[ V= \frac{1}{4\pi\varepsilon_0} \frac{(+q)}{R} \]
Step 4: Compute work done. Charge brought: \[ q_0=-q \] Hence \[ W=q_0V \] \[ W=(-q) \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{R} \right) \] \[ W= \frac{1}{4\pi\varepsilon_0} \left( \frac{-q^2}{R} \right) \] Thus, \[ \boxed{ W= \frac{1}{4\pi\varepsilon_0} \left( \frac{-q^2}{R} \right) } \] Therefore option (A) is correct.
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